Invertible operator with countable spectrum Let $H$ be  a separable  Hilbert space  and $A$ is  an invertible  bounded operator on $H$. Can we approximate  $A$ with  an invertible operator $B$ such that $sp(B)$ is  a countable set?
Motivation:
If the answer is yes, this  would give's us  an alternative  proof  of  connected ness of $GL(H)$. This  alternative proof is identical to a short and interesting proof  of  connectedness  of  $GL(n,\mathbb{C})$, in page  19 of "Introduction to the  Baums  Connes conjecture" by Alain Valette
 A: The question has already been answered above (by Mike Jury). Here is another way of arguing: Any operator with countable spectrum is in the closure of the invertible operators. So are their limits. But again, the unilateral shift provides an example of an operator not in the closure of the invertibles. For if $\|S-X\|<1$ then $\|I-S^*X\|<1$, and so $S^*X$ is invertible. $X$ cannot be invertible, since otherwise $S^*$ would also be invertible. So no translate $S+\lambda I$ can be approximated by operators with countable spectrum. 
A: I think the answer is "no"; here is a sketch of an argument though I will have to go back and check the details. First, there is nothing special about invertibility here; if every invertible operator is approximable by operators with countable spectrum, then every operator is, just by translation. I claim the unilateral shift $S$ should not be approximable, because it is a Fredholm operator of nonzero index. The set of Fredholm operators is open, and the index is a continuous function on this set, so any sufficiently good approximant would have to be Fredholm of index -1. But I think that if a Fredholm operator has countable spectrum, then the index must be 0 (reason: if 0 is not isolated in the spectrum, then the range won't be closed, and if 0 is isolated (or the operator is invertible) then the index is clearly 0). 
EDIT: Here is a proof that a Fredholm operator $T$ with countable spectrum must have index 0. We may clearly suppose $T$ is not invertible, so 0 is in the spectrum. The first claim is that 0 must be isolated--if not, then since the spectrum is a countable, compact set, 0 must be the limit of a sequence of isolated points in the spectrum. But each isolated point must be an eigenvalue, which means that $T$ would then have a sequence of eigenvalues tending to 0, and would not have closed range and not be Fredholm. So, 0 is isolated. But then $T+\epsilon I$ is invertible for all small $\epsilon$, and by the continuity of the index we have that $ind(T)=0$. 
A: It is well-known that you can approximate a selfadjoint operator $A$ with an operator $B=A+V$ such that $B$ has pure point spectrum. The selfadjoint operator $V$ can have arbitrarily small Hilbert-Schmidt norm.
This is a theorem of Weyl-von Neumann and can be found e.g. in Kato.
But I don't think this works for the nonselfadjoint case. 
EDIT: Christian Remling mentioned and is right, of course, that the point spectrum of $B$ has to be dense in the continuous spectrum of $A$ since the essential spectrum is conserved. 
So the pure point spectrum  of $B$ don't have to be countable.
